Question

give answers of all questions plz
who will give all the ans correct i will mark it as brainliest

Answer 1

(12)
Perimeter of rectangle:- 56 m
difference b/w it's length and breadth:- 12 m
Now,
Let length of rectangle is x m
and breadth of rectangle is y m

As/que

x-y :- 12 m (1)

and,

2(x+y) :- 56 m
(x+y) :- 28 m (2)

Adding eqn (1) & (2)

(x-y) + (x+y) = (12+28)m
x-y+x+y = 40 m
2x = 40 m
x = 20 m

Now,
Length :- 20 m
Breadth :- (20-12)m
Breadth :- 8 m

Area of rectangle = l×b
= (20×8)m
= 160 m

(13)

let length of original rect. :- x m
and breadth :- y m
then,
area of original rect. :- (xy)sq.m

Now,
length of new rect. :- (3x)m
and breadth :- (3y)m
then,
area of new rect. :-
(9xy)sq.m

Now,
ratio of areas of original rect. to new rect. is given by

= area of original rect./area of new rect.

= (xy)sq.m/(9xy)sq.m
= (1/9)
= (1:9)

(14).
length :- 25 m
breadth :- 18 m
Now,
= breadth of both path :- 3m

= length of path parallel to length of rectangular field :- (25-3)m
:- 22 m
= length of path parallel to breadth of rectangular field :- (18-3)m
:- 15 m
= area of both path :- [(22×3)+(15×3)+(3×3)]sq.m
= (66+45+9)sq.m
= (120)sq.m
Now,

= rate of paving the path is Rs 12/sq.m
= total cost of paving of path :- Rs(12×120)
:- Rs 1440